On k traid sequences

Vol. 8 No. 4 (1985) 799-804

ON

k-TRIAD

SEQUENCES

HANSRAJ GUPTA

Panjab University, Chandigarh, India 160014 and

K. SlNGH

Department of Mathematics and Statistics

University of New Brunswick, Fredericton, Canada E3B 5A3 (Received May 29, 1984 and in revised June 29, 1984)

ABSTRACT. At the conference of the Indian Mathematical Society held at Allahabad

in December 1981, S. P. Mohanty and A. M. S. Ramasamy pointed out that the three numbers I, 2, 7, have the following property: the product of any two of them increased by 2 is a perfect square. They then showed that there is no fourth integer which hsares this property with all of them. They used Pell’s equation and the theory of quadratic residues to prove their statement. In this paper, we show that their statement holds for a very large set of triads and our proof of the statement is very simple.

KEY WORDS AND PHPASES. Pell’s equation, congruence, Fibonacci, sequence.

1980 AMS SUBJECT CLASSIFICATION CODE. 10A10.

i. INTRODUCTION.

DEFINITION. Given any integer k, three numbers a_I,

a2,

a

3 are said to form a k-triad, if the numbers

ala

2

+

k,

ala

3

+

k and

a2a

3

+

k (I.I)

are all perfect squares.

An ascending sequence of integers a

I, a2, a3, an

is said to be a k-triad sequence if every three consecutive elements of the sequence form a k-triad.

Evidently, if (I.i) is a k-triad sequence, then

a2,

a3,

a

4 an (1.2)

is also a k-triad sequence.

2. CONSTRUCTION OF SEQUENCES (I.i).

In what follows, small letters denote integers and

c.’s

are positive.

i

Given any integer k, we can choose two integers a

I and

a2,

a2 >

al,

such that

ala

2

+

k is a perfect square: 2

ala

2

+

k cI

The problem of constructing the sequence (I.I) then reduces to finding a number a

3 such that both

ala

3

+

k and

a2a

3

+

k will be squares.

2 2

Let

ala

3

+

k x

a2a

3

+

k y

Set x a

+

Cl,

y a

2

+

cI

Then from (2.2) we have (a

2

al)a

3 (y x)(y

+

x) (a

2

-al)(a

I

+

2cI

+

a2);

so that a

3 aI

+

a2

+

2cI

We assert that this value of a

3 actually satisfies our requirements. In fact, we have

Similarly

ala

3

+

k

al(a

I

+

2cI

+

a

2)

+

k

2 a

+

2alc

I

+

(ala

2

+

k)

2 2

a +

2alc

+

c

I

(a

+

ci)2

+

k (a

2

+

Cl

)2

a2a

3 2

Writing c

2

a2a

3

+

k, we have c

2 a2

+

cI

Notice that (2.3) provides a formula for writing the third element of a k-triad sequence interms of

al,

a

2 and c

I.

Applying this procedure to (1.2), we get a

4 a2 + a3 + 2c2 a

2

+

(aI

+

a2

+

2c

I)

+

2(a2

+

c

I)

a + 4a

2

+

4c

Treating this as a formula for the fourth element of a k-triad sequence and applying it to (1.2), we get

a

5 a2

+

4a3

+

4c2 4a

I

+

9a2

+

12cI

The process can be repeated until the desired number of elements of

(I.I)

has been obtained.

(2.1)

(2.2)

(2.3)

(2.4)

(2.5)

(2.6)

Then assuming that

we obtain

aj

uja

I

+

vja

2

+

wjc

I

aj

uja

2

+

vja

3

+

wjc

2

uja

2

+

vj(a

I

+

a2

+

2c

I)

+

wj(a

2

+

c

I)

vja

+

(uj

+

vj

+

wj)a

2

+

(2vj

+

wj)c

This gives us the following recurrence relations:

uj+

I vj, vj+ u.

+

v. + Wo,

wj

+I Identically:

uj+

I 0 0 u.

vj+

I I I v

wj+

0 2 I

wj

or

uj

1 -I

uj+

1

vj

i

0 0

jv

+I

wj

0

LWj+l

Using the same technique, from (2.5), we obtain 2v.

+w.

In general,

Assuming that c

3 a3 + c2 (a

I

+

a2

+

2c

I)

+

(a2

+

c

I)

a

+

2a

2

+

3c

c. =a. +c.

j-I

cj

rja

I

+

sja

2

+

tjc

I

there is no difficulty in obtaining the recurrence relations: r

+I 0 I 0

I

rl

s

+I I I s

t 0 2 it_

+I

and

rj

1 1 -I

rj+

1

sj

0 0

sj+

tj

2 0 I

tj+

While relations (2.9) and (2.12) enable us to find expressions for

ao’S

and

c.’s

as takes value in ascending order of magnitude, the relations (2.10) and (2.13) enable us to extend them in the opposite direction.

(2.9)

(2.10)

(2.11)

(2.12)

Our k-triad sequence can thus be defined for all integral values of the subscript; therefore, the sequence of

c.’s

is defined for all integral values of the subscript.

3. IDENTIFICATION OF THE COEFFICIENTS u, v, w AND r, s, t.

We hardly would expect that the Fibonacci sequence has anything to do with the coefficients u, v, w and r, s, t introduced in the preceding section. But the unexpected happens.

Recall that the Fibonacci sequence is defined by the recurrence relations:

f0

0

fl

I,

f f

+

m > 2

m m-2

fm-i

This definiton can be extended to negative integral subscripts by noting that

m+l

f (-i) f

-m m

From results (2.3), (2.6) and

(2.7),

it will be seen that for the values 3, 4 and 5;

aj

f2

2

al

+

f2.

j-1

a2

+

2fj

-2 fj-1 cI

so that for these values of j,

uj

f2.

j-2 v

f2.

j-I

wj

2fj_2 fj-I

From (2.9), we will now have

2

f2.

f2.

Uj+l

fj-I

Vj+l

j-2

+

-i

+

2fj

-2

fj

-I

(fj-2

+

fj-i )2

f2.

and

wj+

2f

2.

fj_

j-I

+

2fj-2 i

2f.j_l

(fj-i

+

fj-2

2fj_l f’j

(3.1)

(3.2)

We now leave it to the reader to complete the induction and show that

aj

f2

j-2

al

+

f2.

]-I

a2

+

2fj

-2

fj

-i ci

for all integral values of j.

(3.3)

It is no more difficult to prove that for all integral values of

cj

fj_2fj_lal

+

fj_ifja

2

+

(f2

_j

fj-2fj-I )el

4. S! OF

CONSECUTIVE a’s.

From

(2.11),

we have

cj

cj_

1

a.

Hence we have an

interesting

summation formula n

a. c c n > m

j=m+l n m

This provides a good check on the values of

a.’s

and

c.’s.

5. A

GENERALIZATION

OF THE STATEMENT OF MOHANTY AND RAMASAMY.

Our _{generalization can be stated in the form}

of the _following.

THEOREM. If

al, a2,

a3

is a k-triad and k 2 (mod 4), then there is no

integer a for which

(3.4)

ala

+

k,

a2a

+

k,

a3a

+ k are all perfect squares.

We need two lemmas for the proof of our theorem.

LEMMA I. Only two _{of the three numbers a I, a2, a3} are odd. PROOF. If a and a

2 are both even, let 2

ala

2

+

k cI

This implies that c

I is even. Modulo 4, we have

2 0 (mod 4)

This is impossible. Hence both a

I and a2 cannot be even. If a and a are of opposite parity, then since

2 a

3 a + a2

+

2cI

a

3 must be odd, and our lemma holds. If a and a

2 are both odd, then a3 is even. This completes the proof of our lemma.

LEMMA 2. The difference of the two odd elements of the given k-triad is congruent to 2 (mod 4).

PROOF. First let a

I and a2 be the odd elements of the k-triad. Then 0 or 2 (mod 4).

a 2 aI

But a

2 aI cannot be congruent to 0 (mod 4). Suppose a2 aI

H 0 (mod 4) and let

a

2 a

+

4d for some integer d. 2

Since

ala

2

+

k cI cI must be odd.

As

ala

₂

al(a

_I

+

4d) E

a

I (mod 4)

we will have

I

+

2

c

I (mod 4).

This is impossible. Hence a

2 aI 2 (mod 4)

Next let a

I and a3 be the two odd elements of the k-triad. Then a2 is necessarily 2

even. Again since

ala

2

+

k cI cI must be even. We must, therefore, have

a

2

+

2 0 (mod 4)

Hence a

2 2 (mod 4)

Now a

3 a

+

a2

+

2c and c is even.

Hence a

3 aI 2 (mod 4).

The case in which a

2 and a3 are the odd elements, can be dealt with in the same manner and the lemma is proven.

PROOF OF THE THEOREM. Since the existence or non-existence of the number a does not depend on the order in which the three expressions are taken, we can assume that a

For some positive integers x, y, z, let 2

ala

+

k x (i)

2

a2a

+

k y (ii)

2

a3a

+

k z (iii)

From (i) it is evident that x is even. Replace x by 2g, a (which is even)

by 2h, and k (which is congruent to 2 (mod 4)) by 2q where q is odd. Then,

(i) takes the form

2

ha

+

q (2g) (iv)

Since q is odd and the right-hand side is even, h and a must both be odd. This implies that y and z are both odd. Now substracting (ii) and (iii),

we have

2 2

0 (mod 4). (v)

(a

3

a2)a

z y

Since a

3 a2 2 (mod 4), (v) implies that a is even. This contradicts the earlier statement that a is odd. Hence a does not exis and we are through.

-4 -3 -2 -I 0 i 2 3 4 5 6

a. -67 -25 -i0 -3 -i 2 5 15 38 I01 263 690

c. 41 16 6 3 2 4 9 24 62 163 426 1116

f. -3 2 -I I 0 I i 2 3 5 8 13

u

64 25 9 4 i I 0 i I 4 9 25

v

25 9 4 i I 0 i I 4 9 25 64

w. -80 -30 -12 -4 -2 0 0 2 4 12 30 80

r. -40 -15 -6 -2 -I 0 0 i 2 6 15 40

s. -15 -6 -2 -i 0 0 I 2 6 15 40 104

t. 49 19 7 3 i i i 3 7 19 49 129

k 6, a

I 2, a2 5, cI 4

ACKNOWLEDGEMENT. The second author was partially funded by a travel grant from the University of New Brunswick.

REFERENCE